Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k + 1}{k^2 + 7k + 6} \times \dfrac{k^2 + k - 72}{k - 8} $
Answer: First factor out any common factors. $x = \dfrac{k + 1}{k^2 + 7k + 6} \times \dfrac{k^2 + k - 72}{k - 8} $ Then factor the quadratic expressions. $x = \dfrac {k + 1} {(k + 1)(k + 6)} \times \dfrac {(k - 8)(k + 9)} {k - 8} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(k + 1) \times (k - 8)(k + 9) } { (k + 1)(k + 6) \times (k - 8)} $ $x = \dfrac {(k - 8)(k + 9)(k + 1)} {(k + 1)(k + 6)(k - 8)} $ Notice that $(k + 1)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {(k - 8)(k + 9)\cancel{(k + 1)}} {\cancel{(k + 1)}(k + 6)(k - 8)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $x = \dfrac {\cancel{(k - 8)}(k + 9)\cancel{(k + 1)}} {\cancel{(k + 1)}(k + 6)\cancel{(k - 8)}} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $x = \dfrac {k + 9} {k + 6} $ $ x = \dfrac{k + 9}{k + 6}; k \neq -1; k \neq 8 $